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(C) For $f(x)$ to be differentiable everywhere,it must be continuous and differentiable at $x = 1$ and $x = -1$. Since $f(x)$ is an even function,we c

Vedclass · Accepted Answer

$(\frac{1}{2}, \frac{3}{2})$

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(C) For $f(x)$ to be differentiable everywhere,it must be continuous and differentiable at $x = 1$ and $x = -1$. Since $f(x)$ is an even function,we check $x = 1$.Continuity at $x = 1$: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$.$\lim_{x 	o 1^-} (\alpha x^2 - \beta) = \alpha - \beta$.$\lim_{x 	o 1^+} (-\frac{1}{x}) = -1$.So,$\alpha - \beta = -1$  (Equation $1$).Differentiability at $x = 1$: $f'(1^-) = f'(1^+)$.$f'(x) = 2\alpha x$ for $|x|  1$.$f'(1^-) = 2\alpha(1) = 2\alpha$.$f'(1^+) = \frac{1}{(1)^2} = 1$.Thus,$2\alpha = 1 \implies \alpha = \frac{1}{2}$.Substituting $\alpha = \frac{1}{2}$ into Equation $1$: $\frac{1}{2} - \beta = -1 \implies \beta = \frac{3}{2}$.Therefore,the ordered pair $(\alpha, \beta) = (\frac{1}{2}, \frac{3}{2})$.

If $\alpha$ and $\beta$ are such that the function $f(x)$ defined by $f(x) = \begin{cases} \alpha x^2 - \beta, & |x| < 1 \\ \frac{-1}{|x|}, & |x| \ge 1 \end{cases}$ is differentiable everywhere,then the ordered pair $(\alpha, \beta) =$

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